1. Quick question

If $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)$

What is $\left$?

If $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right)$

What is $\left$?
so square both sides of your initial equation
$(x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2$

and evaluate further if you know what the x1,x2,....etc are

so square both sides of your initial equation
$(x)^2 = \frac{1}{n^2}\left(x_{1}+x_{2}+...+x_{n}\right)^2$

and evaluate further if you know what the x1,x2,....etc are
It's not $\left^2$ I want to find. It's $\left
$
.

Basically I'm trying to solve for $x$ the equation:

$2.61 = \sqrt{\left-\left^2}$ to find the value of $x$, where $2.61$ is the $\text{standard deviation}$ of the data $8,7, 4, 10, 6, x$.

I know that $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right)$.

$\Rightarrow \left = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2$.

So $2.61 = \sqrt{\left-\frac{1}{36}\left(35+x\right)^2}$

But I've no idea how to get $\left$.

It's not $\left^2$ I want to find. It's $\left
$
.

Basically I'm trying to solve for $x$ the equation:

$2.61 = \sqrt{\left-\left^2}$ to find the value of $x$, where $2.61$ is the $\text{standard deviation}$ of the data $8,7, 4, 10, 6, x$.

I know that $\left = \frac{1}{n}\left(x_{1}+x_{2}+...+x_{n}\right) = \frac{1}{6}\left(8+7+4+10+6+x\right) = \frac{1}{6}\left(35+x\right)$.

$\Rightarrow \left = \left[\frac{1}{6}\left(35+x\right)\right]^2 = \frac{1}{36}\left(35+x\right)^2$.

So $2.61 = \sqrt{\left-\frac{1}{36}\left(35+x\right)^2}$

But I've no idea how to get $\left$.
$s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }$.

5. Originally Posted by mr fantastic
$s = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \overline{X} }$.
$\left = \frac{\displaystyle\sum_{i=1}^n x_i^2}{n}?$

6. Just tell me how to calculate, or give me definition of, $\left$ and I will take care of the rest.

Just tell me how to calculate, or give me definition of, $\left$ and I will take care of the rest.
$ = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}$.
$ = \frac{x_1^2 + x_2^2 + x_3^2 + .... + x_n^2}{n}$.